Question

How do you differentiate `f(t)= sqrt(t/2+3) / (-2t-6)` using the quotient rule?

Answer 1

`f'(t)=(t+9)/(2*(2*t+6)^2*sqrt(t/2+3))`

Explanation:

We get

`f'(t)=(1/4*(t/2+3)^(-1/2)*(-2t-6)+sqrt(t/2+3)*2)/(2*t+6)^2`
multiplying numerator and denominator by

`sqrt(t/2+3)`
`f'(t)=(-1/2t-3/2+t+6)/(sqrt(t/2+3)*(2*t+6)^2)`
this is

`f'(t)=(t+9)/(2*(2t+6)^2*sqrt(t/2+3))`